Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q13E

Expert-verified
Essential Calculus: Early Transcendentals
Found in: Page 443
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Determine whether the series is convergent or divergent. If its convergent, find its sum.

\(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{(n - 1)}}}}} \)

The given geometric series is divergent.

See the step by step solution

Step by Step Solution

Rewriting the equation.

\(\begin{aligned}{l}\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {{3^n}{e^{{{(n - 1)}^{ - 1}}}}} \\\end{aligned}\)

\( = \sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}} \)

Writing in terms of b and q.

\(\sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}} \) is a geometric series with first term b = \(\frac{3}{e}\)and common ratio q = \(\frac{3}{e}\).

Applying ratio test.

Since \(\left| q \right| = \left| {\frac{3}{e}} \right| > 1\), then by ratio test the series is divergent.

Hence, \(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} \) is divergent.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.